In the Mayer-Vietoris sequence for X = A ∪ B, the map H_n(A ∩ B) → H_n(A) ⊕ H_n(B) sends [c] to (i_*[c], j_*[c]) where i, j are the inclusions. What does it mean when this map is injective?
AEvery cycle in A ∩ B that becomes trivial in A also becomes trivial in B
BEvery cycle in A ∩ B that becomes trivial in both A and B separately is already trivial in A ∩ B
CA ∩ B is a deformation retract of A
DH_n(X) ≅ H_n(A) ⊕ H_n(B)
Injectivity of the map (i_*, -j_*) means: if a cycle c in A ∩ B becomes a boundary both in A and in B, then c was already a boundary in A ∩ B. Equivalently: no nontrivial cycle of A ∩ B is 'killed' by passing to A and B simultaneously. By exactness, injectivity of this map forces the connecting homomorphism H_n(X) → H_{n-1}(A ∩ B) to be zero, which means the sequence splits into short exact sequences 0 → coker → H_n(X) → ker → 0, simplifying the computation.
Question 2 True / False
Using Mayer-Vietoris with A and B as open hemispheres of S^n, the connecting homomorphism H_n(S^n) → H_{n-1}(S^{n-1}) is an isomorphism for n ≥ 2.
TTrue
FFalse
Answer: True
Both hemispheres are contractible, so H_k(A) = H_k(B) = 0 for k > 0. The Mayer-Vietoris sequence gives: 0 → H_n(S^n) → H_{n-1}(S^{n-1}) → 0 for n ≥ 2, since the terms H_n(A) ⊕ H_n(B) and H_{n-1}(A) ⊕ H_{n-1}(B) vanish. By exactness, the connecting homomorphism is an isomorphism. This is the key step in the inductive computation of H_*(S^n).
Question 3 Short Answer
Compute H_1 of the torus T^2 using Mayer-Vietoris, decomposing T^2 as the union of two open cylinders.
Think about your answer, then reveal below.
Model answer: Let A and B be open neighborhoods of the two half-tori, each homotopy equivalent to S^1 (deformation retracting to a circle). Their intersection A ∩ B is two disjoint open annuli, each homotopy equivalent to S^1, so H_1(A ∩ B) ≅ Z ⊕ Z and H_0(A ∩ B) ≅ Z ⊕ Z. The Mayer-Vietoris sequence gives: H_1(A ∩ B) → H_1(A) ⊕ H_1(B) → H_1(T^2) → H_0(A ∩ B) → H_0(A) ⊕ H_0(B). This is Z^2 → Z^2 → H_1(T^2) → Z^2 → Z^2. Working out the maps: the first map is [1,1; 1,1] (both circles in A ∩ B include to the same generator in each piece), which has kernel Z. Chasing through gives H_1(T^2) ≅ Z^2.
The Mayer-Vietoris computation for the torus is a standard exercise. The key subtlety is tracking the inclusion maps carefully — each component of A ∩ B maps to A and B, and the relationship between these maps determines the kernel and cokernel that produce H_1(T^2). This computation is much more tractable than working directly with singular chains, demonstrating the power of the Mayer-Vietoris sequence.
Question 4 Multiple Choice
The Mayer-Vietoris sequence is the homological analogue of which principle from combinatorics?
Inclusion-exclusion for cardinalities says |A ∪ B| = |A| + |B| - |A ∩ B|. The Mayer-Vietoris sequence is the homological upgrade: it relates H_*(A ∪ B) to H_*(A), H_*(B), and H_*(A ∩ B). The Euler characteristic satisfies the exact analogue: χ(A ∪ B) = χ(A) + χ(B) - χ(A ∩ B), which follows from the Mayer-Vietoris sequence. But the full long exact sequence carries more information than just the alternating sum — it encodes how cycles in the pieces interact.